初中
数学
中等
来源: 教材例题
知识点: 初中数学
答案预览
点击下方'查看答案'按钮查看详细解析并跳转到题目详情页
直接前往详情页
练习完成!
恭喜您完成了本次练习,继续加油提升自己的知识水平!
学习建议
您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":353,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级调查中,某学生记录了全班30名同学的身高情况,并将数据整理成如下频数分布表:\n\n身高区间(cm) | 频数\n---------------|------\n150~155 | 4\n155~160 | 8\n160~165 | 12\n165~170 | 5\n170~175 | 1\n\n请问这组数据的众数所在的区间是哪一个?","answer":"C","explanation":"众数是指一组数据中出现次数最多的数值。在本题中,频数表示每个身高区间内的人数。观察频数分布表可知:150~155有4人,155~160有8人,160~165有12人,165~170有5人,170~175有1人。其中,160~165这一区间的频数最大(12人),因此众数所在的区间是160~165。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:43:05","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"150~155","is_correct":0},{"id":"B","content":"155~160","is_correct":0},{"id":"C","content":"160~165","is_correct":1},{"id":"D","content":"165~170","is_correct":0}]},{"id":2280,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在数轴上,点A表示的数是-5,点B与点A的距离是8个单位长度,且点B在原点右侧。若点C是点A和点B之间的一个点,且AC:CB = 3:1,则点C表示的数是___。","answer":"1","explanation":"首先,点A表示-5,点B在原点右侧且与A距离为8,因此点B表示的数是-5 + 8 = 3。点C在A和B之间,且AC:CB = 3:1,说明点C将线段AB按3:1的比例内分。根据内分点公式,点C的坐标为:(1×(-5) + 3×3) ÷ (3+1) = (-5 + 9) ÷ 4 = 4 ÷ 4 = 1。因此,点C表示的数是1。此题综合考查了数轴上的距离、位置关系以及线段的按比例分割,符合七年级数轴与有理数运算的综合应用要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 16:27:13","updated_at":"2026-01-09 16:27:13","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":131,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个长方形的长比宽多5厘米,若其周长为30厘米,则这个长方形的宽是______厘米。","answer":"5","explanation":"设长方形的宽为x厘米,则长为(x + 5)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入得:2 × (x + x + 5) = 30,即2 × (2x + 5) = 30,化简得4x + 10 = 30,解得4x = 20,x = 5。因此,宽为5厘米。本题结合代数设未知数与一元一次方程求解,符合初一学生对方程和几何基础的学习要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 08:59:12","updated_at":"2025-12-24 08:59:12","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2134,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时,将方程 3x + 5 = 20 的第一步写为 3x = 15。请问该学生在这一步中运用了等式的哪一条基本性质?","answer":"B","explanation":"该学生将方程 3x + 5 = 20 变形为 3x = 15,是将等式两边同时减去了 5,从而消去左边的常数项。这一操作依据的是等式的基本性质:等式两边同时减去同一个数,等式仍然成立。因此正确答案是 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 12:56:39","updated_at":"2026-01-09 12:56:39","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"等式两边同时加上同一个数,等式仍然成立","is_correct":0},{"id":"B","content":"等式两边同时减去同一个数,等式仍然成立","is_correct":1},{"id":"C","content":"等式两边同时乘以同一个数,等式仍然成立","is_correct":0},{"id":"D","content":"等式两边同时除以同一个数,等式仍然成立","is_correct":0}]},{"id":574,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间时,记录了5位同学一周内每天阅读的分钟数,分别为:25、30、35、40、45。如果这5位同学每天阅读时间都增加10分钟,那么他们新的平均阅读时间是多少分钟?","answer":"C","explanation":"首先计算原始数据的平均阅读时间:(25 + 30 + 35 + 40 + 45) ÷ 5 = 175 ÷ 5 = 35(分钟)。每位同学的阅读时间都增加10分钟,相当于整体平均数也增加10分钟。因此新的平均阅读时间为:35 + 10 = 45(分钟)。本题考查数据的整理与描述中的平均数概念,属于简单难度,符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:55:35","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"35","is_correct":0},{"id":"B","content":"40","is_correct":0},{"id":"C","content":"45","is_correct":1},{"id":"D","content":"50","is_correct":0}]},{"id":346,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间数据时,记录了10名同学每周阅读的小时数分别为:3,5,4,6,3,7,5,4,5,6。这组数据的众数是多少?","answer":"C","explanation":"众数是指一组数据中出现次数最多的数。将数据从小到大排列为:3,3,4,4,5,5,5,6,6,7。其中,3出现2次,4出现2次,5出现3次,6出现2次,7出现1次。因此,出现次数最多的数是5,所以这组数据的众数是5。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:41:14","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"3","is_correct":0},{"id":"B","content":"4","is_correct":0},{"id":"C","content":"5","is_correct":1},{"id":"D","content":"6","is_correct":0}]},{"id":1362,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生进行校园绿化活动,计划在校园内的一块矩形空地上种植花草。已知这块空地的长比宽多6米,且其周长为44米。为了合理规划种植区域,学校决定在空地内部铺设一条宽度相同的环形步道,步道的内侧形成一个较小的矩形种植区。若铺设步道后,剩余种植区的面积是原空地面积的一半,求步道的宽度。","answer":"设原矩形空地的宽为x米,则长为(x + 6)米。\n根据周长公式:2(长 + 宽) = 44\n代入得:2(x + x + 6) = 44\n化简:2(2x + 6) = 44 → 4x + 12 = 44 → 4x = 32 → x = 8\n所以,原空地的宽为8米,长为8 + 6 = 14米。\n原面积为:8 × 14 = 112平方米。\n设步道的宽度为y米,则内侧种植区的长为(14 - 2y)米,宽为(8 - 2y)米(因为步道在四周,每边减少2y)。\n根据题意,种植区面积是原面积的一半,即:\n(14 - 2y)(8 - 2y) = 112 ÷ 2 = 56\n展开左边:14×8 - 14×2y - 8×2y + 4y² = 56\n即:112 - 28y - 16y + 4y² = 56\n合并同类项:4y² - 44y + 112 = 56\n移项得:4y² - 44y + 56 = 0\n两边同除以4:y² - 11y + 14 = 0\n使用求根公式:y = [11 ± √(121 - 56)] \/ 2 = [11 ± √65] \/ 2\n√65 ≈ 8.06,所以y ≈ (11 ± 8.06)\/2\ny₁ ≈ (11 + 8.06)\/2 ≈ 9.53,y₂ ≈ (11 - 8.06)\/2 ≈ 1.47\n由于原空地宽为8米,步道宽度不能超过4米(否则内侧无种植区),故舍去y ≈ 9.53\n因此,步道的宽度约为1.47米。\n但题目要求精确解,故保留根号形式:\ny = (11 - √65)\/2 (舍去较大根)\n经检验,(11 - √65)\/2 ≈ 1.47,符合实际意义。\n答:步道的宽度为(11 - √65)\/2米。","explanation":"本题综合考查了一元一次方程、整式的加减、实数以及几何图形初步中的矩形面积与周长计算。首先通过周长建立方程求出原矩形的长和宽,属于基础应用;接着引入变量表示步道宽度,利用面积关系建立一元二次方程,涉及整式乘法与化简;最后求解一元二次方程并依据实际意义取舍解,体现了数学建模与实际问题结合的能力。题目难度较高,因需多步推理、代数运算及合理性判断,符合困难级别要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:08:35","updated_at":"2026-01-06 11:08:35","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":548,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的运动项目数据时,绘制了如下扇形统计图。其中表示‘篮球’的扇形圆心角为108度,表示‘足球’的扇形圆心角为90度,表示‘跳绳’的扇形圆心角为72度,其余为‘其他’。如果该班共有40名学生,那么喜欢‘其他’运动项目的学生人数是多少?","answer":"C","explanation":"扇形统计图中,每个扇形的圆心角占整个圆(360度)的比例等于该部分人数占总人数的比例。首先计算已知三个项目的圆心角总和:108 + 90 + 72 = 270度。因此,‘其他’项目对应的圆心角为360 - 270 = 90度。90度占360度的比例为90 ÷ 360 = 1\/4。总人数为40人,所以喜欢‘其他’项目的人数为40 × 1\/4 = 10人。因此正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:05:08","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"6人","is_correct":0},{"id":"B","content":"8人","is_correct":0},{"id":"C","content":"10人","is_correct":1},{"id":"D","content":"12人","is_correct":0}]},{"id":1098,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生收集了可回收垃圾共12.5千克,其中废纸占8.3千克,塑料瓶占2.7千克,其余为金属。若金属的重量用代数式表示为 12.5 - 8.3 - 2.7,则金属的重量是___千克。","answer":"1.5","explanation":"根据题意,金属的重量等于总重量减去废纸和塑料瓶的重量,即 12.5 - 8.3 - 2.7。先计算 12.5 - 8.3 = 4.2,再计算 4.2 - 2.7 = 1.5。因此,金属的重量是1.5千克。本题考查有理数的加减运算在实际问题中的应用,属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:57:01","updated_at":"2026-01-06 08:57:01","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2301,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次班级数学测验中,某学生记录了5名同学完成一次函数应用题所用的时间(单位:分钟),分别为:8、10、12、10、15。这组数据的中位数和众数分别是多少?","answer":"A","explanation":"首先将数据从小到大排列:8、10、10、12、15。共有5个数据,奇数个,因此中位数是中间的那个数,即第3个数,为10。众数是出现次数最多的数,10出现了两次,其余数各出现一次,因此众数是10。所以正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:44:00","updated_at":"2026-01-10 10:44:00","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"中位数是10,众数是10","is_correct":1},{"id":"B","content":"中位数是12,众数是10","is_correct":0},{"id":"C","content":"中位数是10,众数是12","is_correct":0},{"id":"D","content":"中位数是11,众数是10","is_correct":0}]}]